php - 为什么这个变量没有被发布并给出错误？

Question

为什么 bakeryid 变量没有发布在我的表单中？我收到的错误是

"Notice:Undefined variable bakeryid"

我有两个页面，一个显示表单，第二个是表单的操作。第二种形式也一直说它是未定义的。 bakeryid 是每个蛋糕订单的 ID。

$sql = mysqli_query($con,"SELECT `firstname`, `bakeryid`, `order` FROM cakes");

$bakeryid = $_POST['bakeryid'];

?>
  <table border='2'>
    <th>First Name</th>
    <th>Order</th>
<?php
  echo '<form name="display" method="POST" action="cakephp.php">';
  while($row = mysqli_fetch_array($sql))
  {
    echo "<tr>";
    echo "<td>" . $row['firstname'] . "</td>";
    echo "<td>" . $row['order'] . "</td>";
    echo '<td><input type="hidden" name="bakeryid" value="' . $bakeryid . '"/></td>';
    echo '<td><input type="hidden" name="memid" value="' . $memid . '"/><input type="submit" name="takeorder" value="Take Order" ></td>';
    echo "</tr>";
  }
  echo "</form>";
  echo "</table>";

答案:

echo '<td><input type="hidden" name="bakeryid" value="' . $row['bakeryid'] . '"/></td>';

Answer 1

您使用 $_POST['bakeryid'] 设置 $bakeryid，然后使用 $bakeryid 定义 Bakeryid。

请尝试:

$sql = mysqli_query($con,"SELECT `firstname`, `bakeryid`, `order` FROM cakes c INNER JOIN members m ON c.memid = m.memid");

$bakeryid = $_POST['bakeryid']; // this line is unnecessary

    ?> <table border='2'>
            <th>First Name</th>
            <th>Order</th>
<?php
        echo '<form name="display" method="POST" action="cakephp.php">';
while($row = mysqli_fetch_array($sql))
            {
                    echo "<tr>";
                    echo "<td>" . $row['firstname'] . "</td>";
                    echo "<td>" . $row['order'] . "</td>";
                    echo '<td><input type="hidden" name="bakeryid" value="' . $row['bakeryid'] . '"/></td>'; // this line changed
                    echo  '<td><input type="hidden" name="memid" value="' . $memid . '"/><input type="submit" name="takeorder" value="Take Order" ></td>';
                echo "</tr>";
            }
echo "</form>";